forking firecracks i cant take it

sick arab culture fiecracks and gunshots during wedding ceremonials

There are 85 telephones in the editorial department of Glencoe Publishing company. How many 2 way connections can be made among the office phones?

Can anyone help me solve this?

I tried 85 C 2 thinking that order doesn't matter

but the answer i got was wrong

what 'should' the answer be?

oh wait never mind

it should have been 3570

and I got it

yep

if you did distinguish between the caller and receiver, how many would it be instead?

I don't know maybe order would matter and we could use nPr

well, with each two way connection, how many different orders are there?

I don't know how i would do that

I think you're overcomplicating it. If there was a call between us, either you called me or I called you.

So for that two-way connection there's two possible orders.

So in this case the number of ways should double if order matters, i.e. 2*3570=7140.

ya, so would i double the possiblity

ohh ok

But I should also be able to answer it like so: There are 85 choices for the caller, and once I pick them there's 84 choices left for the receiver.

and 85*84=7140

more generally: $_nP_r = n!/(n-r)! = r!\, _nC_r$

in this case $r=2$ so that's just doubling it

ohh ok

thanks for clearing that, Im doing my math homework and I am kinda bad at math so i get confused a lot

np

Considering a deck of 52 cards, How many different 5-card hands can have 5 cards of the same suit?

for this i thought about 13!

or is it 13*12*11*10*9

well, you've also got to pick which suit it is

and then out of the 13 cards in a suit, pick 5 of them

?

How would I do it?

Well, how many suits are there?

4

in total right?

Right. Suppose you pick hearts.

Then there are 13 hearts to pick from, and you need to choose 5 of them without worrying about the order.

ya

then?

Well, how would you compute that?

13C5?

Right.

And then there's 4 choices of suit, so...

so times 4?

OHH

Right. So 4*13C5

Ok I didn't realize the possibility of there being 4 others lol

Right. alternatively, it's 13C5 for each of the 4 suits

so X+X+X+X=4X

one more quick question, if it says to find the value of n in the following

like for example

C(11,8)=C(11,n)

how would i solve it

Well, what's the LHS in terms of factorials?

do i plug in the given in N!/(n-r)!*r!?

LHS?

start there, yeah

left-hand side

how would i compute c(11,8) in calculater?

or just in general like is one of the r and one of them n?

in general

just what the factorials look like

I don't know I'm lost completely

well, n=11 and r=8 right?

so 11 choose 8 is 11!/(8!*3!)

oh thanks for helping me today bro, I'm kinda in a hurry so i have to go, i still don't quite get it doe haha

Alrighr

Chat, don't forget my integral (entirely)

What's that useful for?

@0celo7 The answer is in The Man Who Knew Infinity.

@0celo7 Hardy to Ramanujan: What they (referring to Euler and Jacobi) had in common, what I see in you, is a love of form. It's all through your notebooks. Let me ask you something. Why do you do it, any of this? Ramanujan to Hardy: Because I have to. I see it. Hardy to Ramanujan: Like Euler. Form for its own sake. An art unto itself.

Does anyone have a good demonstration example for why an implication (if..then) is true even if the first part is false?

@Idomathart Do you think I have some great respect for Ramanujan?

To be honest, number theory seems like a collection of party tricks.

@Jeff "If the Riemann hypothesis is true, then 2+2=4." I know that "2+2=4" is true, so there's no way for anyone to ever falsify that statement.

holy shit did you just prove RH

In order for someone to do so, they'd need to 1) show that the Riemann hypothesis is true (very hard), 2) show that nevertheless 2+2 doesn't equal 4 (impossible).

I think he did

because 2+2=4

@Semiclassical Good. Probably too advanced to be simple. But I can use that idea for a better example. (FYI: I"m trying to come up with examples to teach sophmores).

Does that mean I'm immortal now? Cool.

@0celo7 Perhaps you don't because you don't understand his work. I don't comprise all his work either for the simple fact there are things I didn't study yet, but from what I understood I realized that this mind is so profound one needs days, weeks to recover after the meeting with such a profound mathematical wisdom.

Mmkay. I was looking for an example of a premise that was sufficiently 'out there.'

I don't care to understand it!

...sigh. this conversation makes my head hurt.

:D

(Not the logic conversation)

really? i'm busy looking up the riemann hypothesis. i hope that doesn't make my head hurt

@0celo7 respect for him indeed, glorify pr praise him, dunno think so

Depends on which version you mean. The bare statement of it isn't hard to follow, but understanding all the implications thereof is a huge task

@semi what's the bare statement? i'm reading wikipedia and it went right to the most general (requiring me to know what a Zeta function is).

@0celo7 Sure, maybe you don't prefer some really hard stuff, and especially I'm sure of that when I look at the inequality you gave me to solve above. $$\sqrt{|y-z|}\le \sqrt y+\sqrt z$$

You're right, I don't like hard integrals

I also don't see why I should

Where the zeros of the Riemann zeta function are, is the bare version

@Semiclassical Why not just compute $\zeta^{-1}(0)$

doesn't seem to hard

@0celo7 I appreciate your honesty!

@amWhy True.

I'm not being sarcastic...I meant that as simply a friendly suggestion ;-)

What's simple is the idea that a certain function, however unfamiliar, can only vanish at certain points

@amWhy It's welcome, really.

What that function is, is more involved. Understanding why it would matter is an even longer story

For the purposes of an example, something like Goldbach's conjecture is probably simpler

Easy to state, but entirely beyond proof

Does anyone proficient in the Guassian integers know if they can be used to solve this question? In my posted answer you will find some partial work..

@semi now i'm looking up goldbach's :D

@Idomathart I'd really like to be clear, and start afresh. I do not, and never disliked you, nor meant to ridicule your work. The more enthusiasm, the better! I would like you to express that enthusiasm; but try to be enthusiastic and supportive of others' work too. I'm not saying you aren't; I'm just saying that it has struck me in the last few days that your enthusiasm ends where your work ends. But you are right that we hardly know each other...

And I will work hard to keep and open mind...

@Jeff Kitty cats! Luv'm

@amwhy TY... will you be sad to learn that the brown one has died?

Oh, yes, very sad.

But I understand...my Facebook profile shows both my cats Rinah and Shai, even though I lost Rinah 2 years ago :-(

$$\int_0^1 \left(\frac{\log(1+x)}{1+x}-\frac{\log(1-x)}{1+x}\right) \arctan(x) \textrm{d}x=\frac{\pi^3}{192}+\frac{\log(2)}{2}G$$ :D

@Agawa001 lol

Gentlemen, ladies, please. There's an 'ignore' button under everyone's profile - use it, if you can't stand being around each other.

Indeed. Let's have math, not sound and fury

@Idomathart i can solve what is before the annoying arctan

@Agawa001 Indeed, that arctan is pretty annoying and makes things more complicated.

A big hint, I suspect, is what representation of $G$ you end with

i m not expert with trigo-logarithmic integrals

@JasperLoy Be positive in all circumstances, this helps you to get over anything.

→ 31 messages moved to Trashcan

@Semiclassical Yeah. Still you also have $\log(2)/2$ in front.

Aye.

@ArtOfCode these lines will be recycled ?

@Agawa001 quite probably

For reference, $\displaystyle G=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}$

for the purpose of this problem, though, the representation $G=\int_0^1 \arctan{t}\,\frac{dt}{t}$ seems more relevant

though that log(2) is a pain...

@JasperLoy Good night.

@JasperLoy Not that painful if you look for a double integral. The preferred form is $$\int_0^1 \int_0^x f(x,y)\textrm{d}y \textrm{d}x$$

A possible symmetry would lead to

$$\frac{1}{2}\int_0^1 \int_0^1 f(x,y)\textrm{d}y \textrm{d}x$$

hmm, so view the integrand as an indefinite integral

@0celo7 that room is private for a reason, I don't just go giving access to everyone :)

@ArtOfCode It was worth a try!

@JasperLoy i would be happy to receive scolds in my mailbox rather than social-network junks and wind blowing echo, and brownrecluse' cobweb

Now if the expression is separable, it's about having an integral that leads to $\log(2)$ and one that leads to $G$, and considering $1/2$ in front, we get exactly what we want.

Now, $\frac{\pi^3}{192}$ would be forced to be the residual result of such a process of creating a double integral as mentioned above such that together to get the initial integral.

Of course, $\pi^3/192$ in the form of an integral.

seeing a representation involving $\int_0^x f(x,y)\,dy$ seems tricky

$\int_0^x f(y)\,dy$ is easy, of course

@ArtOfCode Got it...

@amWhy hm?

got what? Access?

Oh...I thought you were policing the room...anyway...no matter but Thanks?

@amWhy what? I denied your request too :P

I wrote "Got it..." to express my acknowledgement of what happened and why

oh

Does that make sense, or am I misunderstanding the "time out" I received...I thought you dealt it...

Thanks for the fairness

you got a suspension? I didn't even know

huh, so you did

I'm sure I invalidated the flags on that

Brief, but yes...

What to do when a user "badmouths" you (me) to other users, and does so dishonestly? Simply flag, I guess?

Or, ignore, I guess.

@amWhy yep

mod flag if it's not obvious or requires more details

Thanks!

My probability book says the answer to a probability question is 1800/360

how

@0celo7 What question? Why would they write 1800/360 when they could have written "5"?

5 is a ridiculous probability too

I completely agree with you!

you have a jar with 6 white chips, 4 black chips, 5 red chips

you draw them and don't replace, what is the probability of getting BBRWW

I got $\frac{4}{15}\cdot\frac{3}{14}\cdot\frac{5}{13}\cdot\frac{6}{12}\cdot\frac{5}{11‌​}=\frac{5}{1001}$

That sure looks "spot on" to me!

well the back of the book says the answer is $1800/360, 360$

Have you checked on-line for any "errata" for the text?

@0celo7 that's just wrong?!

wait a moment

lol

that's not a comma denoting 1800/360 AND 360

it's 360360

the number

and that ratio is indeed what I got

Ahhhh! good for you!

"A man has $n$ keys on his key ring"

only a mathematician could some up with this problem

lol

it's about a drunk guy trying to get into his apartment

@0celo7 Hahahahahaha!

Is this a trick problem?

Is he supposed to pass out before he gets to the third key?

What's the jist of the question? A man has $n$ keys on a key ring, and is drunk...and.....?

He has $n$ keys and doesn't remember which one is which

What is the probability of the third key working on the door

it's $1/n$

Well, you've got a great point about him perhaps passing out! Seriously, since he is "presumed" to be incoherent, then the probability of any choice of key, including the third key, will be 1/n...assuming we have keys that are distinct and numbered 1, 2, ... n

no, it's more subtle than that

to get to the third key you have to go through keys 1 and 2

they're not numbered

they mean, if he picks any three keys, what is the probability the third one will open the door

Ahhh...got it...picking of three keys of n....then the probability of the first of the three not working....then the probability of the second not working, which leaves the third key, which is successful.

this is the "axiomatic" computation

k being the event that key k works

In the end, the probability is equivalent to that of randomly picking a specific key from a set of n, yes?

Yea

but you have to prove that :P

Hello everyone I was asking myself about the following" Does $sum_1^infty ((1/n)-(1/p)) where p are prime numbers converge ? "

@0celo7 Can you understand why the two approaches lead to the correct solution? Why the task at hand amounts to/is equivalent to the task of randomly picking one specific key from n keys?

nope